Which compound forms a green aqueous solution

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

Degger [83]

a) before addition of any KOH :

when we use the Ka equation & Ka = 4 x 10^-8 :

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

= 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

= 9.2 x 10^-5 M

when PH = -㏒[H+]

PH = -㏒(9.2 x 10^-5)

= 4

b)After addition of 25 mL of KOH: this produces a buffer solution

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]

first, we have to get moles of HCIO= molarity * volume

=0.21M * 0.05L

= 0.0105 moles

then, moles of KOH = molarity * volume

= 0.21 * 0.025

=0.00525 moles

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L = 0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

= 0.00525 / 0.075

=0.07 M

and molarity of KCIO = moles KCIO / total volume

= 0.00525 / 0.075

= 0.07 M

and when Ka = 4 x 10^-8

∴Pka =-㏒Ka

= -㏒(4 x 10^-8)

= 7.4

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume

= 0.21 M * 0.05L

= 0.0105 moles

then moles KOH = molarity * volume
= 0.22 M* 0.035 L

=0.0077 moles

∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume

= 8 x 10^-5 / 0.085

= 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

= 0.0077M / 0.085L

= 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

= 0.0105mol / 0.1 L

= 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

= 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO]

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

= 3.8
∴PH = 14- POH

=14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M

V2 is the excess volume added of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

= -㏒0.02

= 1.7

∴PH = 14- POH

= 14- 1.7

= 12.3

8 0

3 years ago

Read 2 more answers

put the contributions to the understanding of the atomic structure in order from most recent at the top to the earliest at the b

r-ruslan [8.4K]

Answer:

From Top to Bottom:

- Democritus coming up with the concept of an atom

- Dalton discovering that atoms are the smallest part of an element

- Rutherford discovering the nucleus of an atom

- Thomson discovering electrons

- Bohr modeling electrons orbiting the nucleus

- Schrodinger modeling electrons in the electron cloud

Explanation:

The best way to think about this is from the inside out. Democrats (who lived long before any of the other scientists mentioned) was the one who thought of the idea of the atom. - Therefore, this must be first because all other choices are elaborations on the idea that atoms exist. Next must be Dalton. Dalton saw atoms as "cannonballs" if you will; a solid mass. So then after that, Rutherford and his gold foil experiment (he discovered that some rays he shot through gold foil were deflected back; ie the existence of concentrated areas in an atom, ie the nucleus). Then we get into the information on electrons. We must start with discovery (Thomson). Heres where it gets complicated. Electrons don't actually orbit the nucleus, they exist in electron clouds. So it would be Bohr, who came up with the idea that electron exist outside the nucleus, then Schrodinger, who elaborated on Bohr's theory. Hope this helps!

Nat, Junior

Accel + AP Chem student

5 0

1 year ago

For the following example, list the given and unknown information (including gratis or moles)

Setler79 [48]

Answer:

9.6 moles O2

Explanation:

I'll assume it is 345 grams, not gratis, of water. Hydrogen's molar mass is 1.01, not 101.

The molar mass of water is 18.0 grams/mole.

Therefore: (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).

The balanced equation states that: 2H20 ⇒ 2H2 +02

It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.

get (1 mole O2/2 moles H2O)*(19.2 moles H2O) or 9.6 moles O2

6 0

2 years ago

The solubility KI is 50 g in 100 g of H2O at 20 °C. if 110 grams of ki are added to 200 grams of H2O ________

Salsk061 [2.6K]

The solubility KI is 50 g in 100 g of H₂O at 20 °C. if 110 grams of ki are added to 200 grams of H₂O the solution will be saturated.

<h3>What is solubility?</h3>

Solubility is a condition where the solute is fully dissolved in the solvent. When fully mixed with the solvent.

Given that 50 g of KI is added to 100 g of water at 20 °C it means 100 g of water can dissolve a maximum of 50 g of KCl.

1 g of water will dissolve an quantity of 0.5 g of KCl.

To assay for 200 g of water: 200 g of water can disintegrate a maximum of (0.5) x 200 g of KCl.

The maximum amount of KCl that will dissolve is 100 g

Actualised amount dissolved = 110 g

when Amount dissolved > Maximum solubility limit

110 g > 100 g

Thus, the solution is saturated.

To learn more about solubility, refer to the below link:

brainly.com/question/8591226

#SPJ4

6 0

1 year ago

Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at th

stepan [7]

Answer:

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

Explanation:

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH= $2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}$

Knowing:

ΔH= 75.5 kJ/mol
$H_{NO}$ = 90.25 kJ/mol
$H_{Cl_{2} }$ = 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
$H_{NOCl}$ =?

Replacing:

75.5 kJ/mol=2* 90.25 kJ/mol + 0 - $H_{NOCl}$

Solving

- $H_{NOCl}$ =75.5 kJ/mol - 2*90.25 kJ/mol

- $H_{NOCl}$ =-105 kJ/mol

$H_{NOCl}$ =105 kJ/mol

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

8 0

2 years ago